Let $f(x) = 4x^{2}-4x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $4x^{2}-4x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 4, b = -4, c = 1$ $ x = \dfrac{+ 4 \pm \sqrt{(-4)^{2} - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$ $ x = \dfrac{4 \pm \sqrt{0}}{8}$ $ x = \dfrac{4 \pm 0}{8}$ $x =\frac{1}{2}$